Dimension of the kernel of a square matrix.

Question

If I have an $n$ by $n$ matrix. Is the dimension of the kernel always equal to the number of zero rows when a matrix is in rref form? I believe it is but if so why is this the case and what would be a proof? Thanks.

Answer 1

Rank of the matrix in RREF form is equal to the number of pivot column, which is equal to the number of non-zero rows.

Nullity of the matrix is equal to number of column $-$ rank of the matrix.

But number of rows is equal to number of columns for our square matrix.

Hence nullity of the matrix is equal to number of rows $-$ number of non-zero rows, which is the number of zero rows.

Answer 2

$\newcommand{\rref}{\operatorname{rref}}$ In fact, something stronger is true:

If $A$ is a matrix, then $\ker(A)=\ker(\rref(A))$.

Proof:

Write $\rref(A)=EA$, where $E$ is a product of elementary matrices (and thus invertible). Since $E$ is invertible, we have $EAx=0$ iff $Ax=0$. Therefore $$ \ker(A)=\{x \mid Ax=0\}=\{x \mid \rref(A)x=EAx=0\}=\ker(\rref(A)). \tag*{$\square$} $$

The question now follows because the dimension of $\ker(\rref(A))$ is the number of zero rows of $\rref(A)$.